Question

The moment of inertia of a thin circular disk about an axis passing through its center and perpendicular to its plane is $I$. Then the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is:
$A)\text{ }I$
$B)\text{ }2I$
$C)\text{ }\dfrac{3}{2}I$
$D)\text{ }\dfrac{5}{2}I$

Answer 1

This problem can be solved by finding out the moment of inertia of the disc about an axis lying in the plane of the disc and passing through its center with the help of the perpendicular axis theorem in terms of $I$. Then, using this value we can find out the moment of inertia of the disc about an axis that is shifted parallel from the former axis such that it now touches the edge of the rim, using the parallel axis theorem.

Formula used:
${{I}_{\text{perpendicular}}}={{I}_{1}}+{{I}_{2}}$
${{I}_{\text{parallel}}}={{I}_{1}}+M{{L}^{2}}$

Complete step-by-step answer:
The perpendicular axis theorem states that the moment of inertia of a body about an axis perpendicular to its plane and passing through a point is given by the sum of any two axes lying on the plane of the body that are at right angles to each other and intersect at the point.
Hence, if ${{I}_{\text{perpendicular}}}$ is the moment of inertia about the axis perpendicular to the plane and ${{I}_{1}}$ and ${{I}_{2}}$ are the moments of inertia about the axes lying in the plane and satisfying the above definition, then, by the perpendicular axis theorem,
${{I}_{\text{perpendicular}}}={{I}_{1}}+{{I}_{2}}$ --(1)
The parallel axis theorem states that the moment of inertia of a body about an axis that is shifted parallel from another axis by a distance is given by the sum of the moment of inertia about the axis from which it is shifted and the product of the mass of the body and the square of the shifted distance.
That is, if ${{I}_{1}}$ is the moment of inertia of a body of mass $M$ about an axis, then the moment of inertia ${{I}_{\text{parallel}}}$ of the body about an axis shifted parallel from the first axis by a distance $L$is given by,
${{I}_{\text{parallel}}}={{I}_{1}}+M{{L}^{2}}$ --(2)
Now, let us analyze the question.
Let the mass of the disc be $M$and its radius be $R$.
The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is given by $\dfrac{M{{R}^{2}}}{2}$.
Hence, according to the problem,
$I=\dfrac{M{{R}^{2}}}{2}$
$\therefore M{{R}^{2}}=2I$ ---(3)
Now, since the disk is symmetrical, its moment of inertia about any axis lying in its plane and passing through its center must be equal. Therefore, let us take two such axes that are at right angles to each other.
Let the moment of inertia about the axes be ${{I}_{1}}$ and ${{I}_{2}}$.
Therefore, as explained, ${{I}_{1}}={{I}_{2}}$ --(4)
Now, since they are at right angles to each other, passing through the center and lying in the plane of the disc, we can use the perpendicular axis theorem.
Hence, using (1), we get,
${{I}_{1}}+{{I}_{2}}=I$
$\therefore 2{{I}_{1}}=2{{I}_{2}}=I$ (using (4))
$\therefore {{I}_{1}}={{I}_{2}}=\dfrac{I}{2}$ --(5)
Now, we have found out the moment of inertia of the disk about the axis passing through its center and lying in its plane. Now, we will shift the axis perpendicular but in the plane such that it touches the edge of the rim.
Obviously, the distance of the center from the edge of the rim is nothing but the radius of the disk. So, the axis has to be shifted by a distance $R$.
Let the moment of inertia about this axis be ${{I}_{\text{required}}}$.
Now, using the parallel axis theorem, that is, using (2), we get,
${{I}_{\text{required}}}={{I}_{1}}+M{{R}^{2}}$
$\therefore {{I}_{\text{required}}}=\dfrac{I}{2}+M{{R}^{2}}$ [Using (5)]
$\therefore {{I}_{\text{required}}}=\dfrac{I}{2}+2I$ [Using (3)]
$\therefore {{I}_{\text{required}}}=\dfrac{5I}{2}$
Therefore, the required value of the moment of inertia is $\dfrac{5I}{2}$.
Therefore, the correct option is D) $\dfrac{5I}{2}$.

Note: While using the perpendicular axis theorem, students must be careful that all the three axes, that is, the two axes lying in the place and the axis perpendicular to the plane pass through a common intersection point. Many times students forget this crucial point and make a complete error in the whole question.
Students must have the moments of inertia of some common bodies like a sphere, ring, disk, etc about some common axes in their mind, since it is impossible to solve almost any question of rotational mechanics without their help. Even if for a body, the moment of inertia about a single axis is known by the student the moments of inertia about the other axes can be derived by him using the parallel and perpendicular axis theorems.